Compute the torque experimenced by a magnetic needle in a uniform magnetic field.

Explanation

Torque Acting on a Bar Magnet in Uniform Magnetic Field. : Consider a magnet of length 2l of pole strength `q_(m)` kept in a uniform magnetic field `vec(B)`. Each pole experinences a force of magnitude `q_(m)` B but acts in opposite direction. Therefore, the net force exerted on the magnet is zero, so that there is no translatory motion. These two forces constitute a couple (about midpoint of bar magnet ) which will rotate and try to align in the direction of the magnetic field `vec(B)` .
The force experienced by north pole, `vec(F_(N)) = q_(m) vec(B)`
The force `e^(x)` perienced by south pole, `vec(F_(S)) = - q_(m) vec(B)`
Adding equations (1) and (2), we get the net force acting on the dipole as
`vec(F ) = vec(F_(N)) = vec(0)` This implies , that the net force acting on the dipole is zero, but forms a couple which tends to rotate the ber magnet clockwise (here) in order to align it along `vec(B)`.
The moment of force or torque experienced by north and south pole about point O is
`vec(tau) = vec(ON) xx vec(F_(N)) + vec(OS) xx vec(F_(s))`
`vec(tau) = vec(ON) xx q_(m) vec(B) + vec(OS) xx (-q_(m) vec(B))`,
By using right hand cork screw rule, we coclude that the total torque is pointing into the paper. Since the magnitudes
`|vec(ON)|= |vec(OS)| = l and |q_(m) vec(B)| = |-q_(m) vec(B)|`
The magnitude of total torque about point O
`tau = l xx q_(m) B sin theta + l xx q_(m) B sin theta`
`tau = 2l xx q_(m) B sin theta`
`tau = p_(m) B sin theta " " (thetafore q_(m) xx 2l = p_(m))`
In vector notation, `tau = p_(m) xx vec(B)`